Answer in Brief

**Answer the following question.**

Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions.

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#### Solution

- Consider an object moving in an x-y plane. Let the initial velocity of the object be `vec"u"` at t = 0 and its velocity at time t be `vec"v"`
- As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal.

`vec"a"_"av" = (vec"v"_2 - vec"v"_1)/("t"_2 - "t"_1) = (("v"_"2x" - "v"_"1x")/("t"_2 - "t"_1))hat"i" + (("v"_"2y" - "v"_"1y")/("t"_2 - "t"_1))`

∴ `vec"a" = ((vec"v" - vec"u"))/(("t - 0"))`

∴ `vec"v" = vec"u" + vec"a""t"` ....(1)

This is the first equation of motion in vector form. - Let the displacement of the object from time t = 0 to t be `vec "s"`

For constant acceleration, `vec"v"_"av" = (vec"v" + vec"u")/2`

`vec"s" = (vec"v"_"av")"t" = ((vec"v" + vec"u")/2)"t" = ((vec"u" + vec"u" + vec"a""t")/2)"t"`

∴ `vec"s" = vec"u""t" + 1/2 vec"a""t"^2` ...(2)

This is the second equation of motion in vector form. - Equations (1) and (2) can be resolved into their x and y components so as to get corresponding scalar equations as follows.

`"v"_"x" = "u"_"x" + "a"_"x" "t"` ....(3)

`"v"_"y" = "u"_"y" + "a"_"y" "t"` ....(4)

`"s"_"x" = "u"_"x" "t" + 1/2 "a"_"x""t"^2` ...(5)

`"s"_"y" = "u"_"y""t" + 1/2"a"_"y""t"^2` ....(6) - It can be seen that equations (3) and (5) involve only the x components of displacement, velocity and acceleration while equations (4) and (6) involve only the y components of these quantities.
- Thus, the motion along the x-direction of the object is completely controlled by the x components of velocity and acceleration while that along the y-direction is completely controlled by the y components of these quantities.
- This shows that the two sets of equations are independent of each other and can be solved independently.

Is there an error in this question or solution?

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